Results 1 to 7 of 7

Thread: Fibonacci Look-a-like

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10

    Fibonacci Look-a-like

    Define a fibonacci series generated by {a,b} such as,
    $\displaystyle u_1=a$
    $\displaystyle u_2=b$
    $\displaystyle u_{n+2}=u_{n+1}+u_{n}$
    (The original fibonacci series is generated by {1,1})
    ($\displaystyle a\not =0$)

    Then there are a number of interesting properties:
    1)If $\displaystyle K$ is the finite continued fraction for $\displaystyle b/a$
    then $\displaystyle \frac{u_{n+1}}{u_n}=[1;1,1...,1,K]$
    Where the $\displaystyle 1$ appears $\displaystyle n-2$ times.

    2)Thus, from here we have that $\displaystyle \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\psi$
    (Thus, any fibonacci series generated by any two real numbers converges to the divine proportion).

    3)The formula for $\displaystyle u_n$ is given by
    $\displaystyle u_n=F(n-2)a+F(n-1)b$
    where $\displaystyle F(n)$ is the n-th fibonacci number.
    But by Binet's formula we have that,
    for the n-th fibonacci number we can find a formula for $\displaystyle u_n$ but it is rather messy and will be omitted. Giving a second method for proving statement 2.
    Last edited by ThePerfectHacker; Nov 22nd 2006 at 09:53 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2005
    Posts
    43
    I think you mean $\displaystyle \phi$ and not $\displaystyle \psi$. Very interesting, though.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Apr 2006
    Posts
    287
    Quote Originally Posted by Treadstone 71
    I think you mean $\displaystyle \phi$ and not $\displaystyle \psi$. Very interesting, though.

    $\displaystyle
    \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi
    $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Natasha1
    $\displaystyle
    \lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=\phi
    $
    That is not important
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    The proof of the second statement is rather neat.

    $\displaystyle 2)\;\;\lim_{n\to\infty}\frac{u_{n+1}}{u_n}\:=\: \phi $

    The limit of the ratio of consecutive Fibonacci numbers is already established.

    . . . . . $\displaystyle [1]\;\lim_{n\to\infty}\frac{F_{n+1}}{F_n}\:=\:\phi \qquad\qquad[2]\;\lim_{n\to\infty}\frac{F_n}{F_{n+1}} \:=\:\frac{1}{\phi}$


    We have: $\displaystyle u_n\:=\:a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)$

    The ratio is: $\displaystyle R\;=\;\frac{u_{n+1}}{u_n}\,=\,\frac{a\!\cdot\!F(n-1) + b\!\cdot\!F(n)}{a\!\cdot\!F(n-2) + b\!\cdot\!F(n-1)} $

    Divide top and bottom by $\displaystyle F_{n-1}:\;\;R\;=\;\frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}$

    Take the limit: $\displaystyle \lim_{n\to\infty}R\;=\;\lim_{n\to\infty}\left( \frac{a + b\!\cdot\!\frac{F(n)}{F_{n-1)}}}{a\!\cdot\!\frac{F(n-2)}{F(n-1)} + b}\right)$

    From [1] amd [2], this becomes: $\displaystyle \frac{a + b\!\cdot\!\phi}{a\!\cdot\!\frac{1}{\phi} + b}$

    Multiply top and bottom by $\displaystyle \phi:\;\;\frac{\phi(a + b\!\cdot\!\phi)}{a + b\!\cdot\!\phi}\;=\;\phi$

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Soroban
    The proof of the second statement is rather neat.

    [size=3]
    My proof is simpler.
    Because of condition 1), since this countinued fractions converges to $\displaystyle [1:1,1,...]=\phi$.
    Last edited by ThePerfectHacker; Jun 4th 2006 at 05:40 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, ThePerfectHacker!

    My proof is simpler.
    Of course it is . . . I wasn't claiming otherwise.

    I was demonstrating a straight algebraic approach to the limit
    . . for those not familiar with continued fractions.

    Please understand: I never ever compete with other posters here.
    I may post an alternate approach or even an improvement
    . . . but never with a snicker or a sneer, implied or otherwise.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fibonacci
    Posted in the Math Puzzles Forum
    Replies: 6
    Last Post: May 15th 2011, 11:17 PM
  2. fibonacci
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: Feb 4th 2010, 01:11 PM
  3. Fibonacci
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 15th 2009, 03:50 PM
  4. Fibonacci
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 25th 2009, 01:31 PM
  5. Fibonacci
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Dec 4th 2008, 06:04 PM

Search Tags


/mathhelpforum @mathhelpforum